A wire fixed at the upper end stretches by length l by applying a force F . The work done in stretching is: - Physics MCQ AIEEE 2004 2026 | ExamTest.live

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A wire fixed at the upper end stretches by length l by applying a force F . The work done in stretching is:

Solution & Step-by-step Explanation

The work done in stretching a wire (elastic potential energy) is given by: W = \frac{1}{2} \times Force \times Extension W = \frac{1}{2} \times F \times l = \frac{Fl}{2} Alternatively, \int F (x) d x = \int k x d x = \frac{1}{2} k l^{2} . Since F = k l at the end, W = \frac{1}{2} (k l) l = \frac{Fl}{2} .

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A wire fixed at the upper end stretches by length l by applying a force F . The work done in stretching is:

F /2 l

Fl

2 Fl

Fl /2

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